Problem: The grades on a geometry midterm at Springer are normally distributed with $\mu = 68$ and $\sigma = 5.0$. Nadia earned a n $81$ on the exam. Find the z-score for Nadia's exam grade. Round to two decimal places.
Explanation: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Nadia's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{81 - {68}}{{5.0}}} $ ${ z \approx 2.60}$ The z-score is $2.60$. In other words, Nadia's score was $2.60$ standard deviations above the mean.